What does the additive subgroup of $\mathbb{R}^3$ generated by the vertices of a regular polyhedron look like? Let $\{v_1,\dots,v_n\} \subset \mathbb{R}^3$ be the vertices of one of the regular polyhedra centered at the origin and let $G = \sum_{k=1}^n v_k \mathbb{Z}$, the additive subgroup of $\mathbb{R}^3$ generated by these vertices.
In some cases, like for a tetrahedron, cube, or octahedron, $G$ is a lattice. Is it always the case or are there some regular polyhedra that will yield a group dense-in-itself?
Clearly if one of these two results is true for a polyhedron, then it also applies to its dual. I don't have any intuition of the answer for the regular dodecahedron or icosahedron though, and even less for the non-convex regular polyhedra.
 A: Let's first consider the corresponding question in the plane.  Consider the subgroup of $\mathbb{R}^2$ generated by the vertices of a regular $n$-gon centered at the origin.  Scaling and rotating, we may assume these vertices are the $n$th roots of unity when we identify $\mathbb{R}^2$ with $\mathbb{C}$, so the subgroup in question is just the subring $\mathbb{Z}[\zeta]\subset\mathbb{C}$ where $\zeta$ is a primitive $n$th root of unity.  By the theory of cyclotomic polynomials, the minimal polynomial of $\zeta$ over $\mathbb{Q}$ has degree $\varphi(n)$ (where $\varphi$ is the totient function), so $\mathbb{Z}[\zeta]$ has rank $\varphi(n)$ as a $\mathbb{Z}$-module.  In particular, as long as $n\neq 1,2,3,4,6$, $\varphi(n)>2$ and so $\mathbb{Z}[\zeta]$ must be dense in $\mathbb{C}$.
It follows that if you take the vertices of a regular $n$-gon in $\mathbb{R}^3$ for $n\neq 1,2,3,4,6$, the subgroup of $\mathbb{R}^3$ they generate will be dense in some plane parallel to the plane of the $n$-gon (namely, the plane you get by scaling away from the origin a factor of $n$, so that the subgroup will contain the center of the scaled $n$-gon as well as all its vertices and thus will contain the subgroup generated by those scaled vertices when you treat the center as the origin).  If you are taking the vertices of a polyhedron that has two non-parallel regular $n$-gons for $n\neq 1,2,3,4,6$ as faces, it then follows that the subgroup its vertices generate will be dense in all of $\mathbb{R}^3$.
